Construct AP = 2 AC.
On the perpendicular to AC from A, construct AO = AC/2.
Circle centered in O through P interesects this perpendicular in M and M'.
(AM+AM')/2 = AO = a/2 and AM.AM' = AP² = 4a².AM and AM' are then the solutions of the quadratic.
Circle centered in C with radius AM and circle centered in A through P intersect in the searched point B.
The choosen solution from the four possible being that with angle A acute, and of vourse the symmetric.
The solution with obtuse A results into C plus 180° (line angles modulo π)
instead of inside angle to triangle.

Some hyperbola...

The locus of points B with angle BCA = 2 BAC is an hyperbola.
The previous construction gives the intersection of this hyperbola
with circle centered in A and radius 2 AC.
The proof comes directly from the previous trapezoid.
The distance from B to perpendicular bisector of AC is then BC/2.
The locus of B is then hyperbola with focus C, directrix line the perpendicular bisector of AC
and excentricity BC/BH = 2.We deduce A is a vertex of this hyperbola, the other one being S with AS = 2/3 AC.
The center is I with AI = 1/3 AC, the other focus is F with AF = -1/3 AC.
The asymptotes are at 60°.